Which option below describes the main difference between analog and digital technology?

What term describes the rate at which charge passes a point in a circuit

docker41 [41]

That's electric "current", usually described in 'Amperes'.

1 Ampere = 1 Coulomb per second ... "the rate at which
charge passes a point in the circuit".

8 0

3 years ago

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The state of strain at a point is plane strain with εx = ε0, εy = –2ε0, γxy = 0, where ε0 is a positive constant. What is the no

Marat540 [252]

Answer:

The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2

Explanation:

Given that

$\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?$

From equation of normal strain in x direction:

$\epsilon_{x_{1}}=\epsilon_{x}cos^{2}\theta+\epsilon_{y}sin^{2}\theta+\gamma_{xy{ sin\theta cos\theta$

Substituting the values:

$\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}$

6 0

3 years ago

How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),

fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice = 37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂ .

Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0

3 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$